Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 4.2
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Exercise 65 Page 188

Practice makes perfect
a By performing inverse operations until x is isolated, we can solve the inequality.

3x+2≥ x-6
3x≥ x-8
2x≥ -8
x≥ -4

The solution set to the inequality is x≥ -4. Therefore, we should shade everything that is greater than or equal to -4. Note that this is a non-strict inequality which means we will keep the endpoint closed.

b To solve the inequality we must first determine its boundary point(s). To do that, we will treat the inequality as if it was an equation and solve for x.

Inequality:& 2x^2-5x<12 Equality:& 2x^2-5x=12 Let's solve it by using the Quadratic Formula.

2x^2-5x=12
2x^2-5x-12=0
x=-( -5)±sqrt(( -5)^2-4( 2)( -12))/2( 2)
â–¼
Evaluate right-hand side
x=5±sqrt((-5)^2-4(2)(-12))/2(2)
x=5±sqrt(25+96)/4
x=5±sqrt(121)/4
x=5±11/4
lcx=.(5-11) /4. & (I) x=.(5+11) /4. & (II)

(I), (II): Add and subtract terms

lx=.-6 /4. x=.16 /4.
lx_1=-1.5 x_2=4

The boundaries of the inequality is x=-1.5 and =4. Let's mark these solutions on a number line. We will keep the points open since this is a strict inequality. To identify where we should shade the number line, we will also include a few test points.

By substituting x with -3, 0, and 5 in the inequality we can determine where the inequality holds true. |c|c|c| [-0.8em] x & 2x^2-5x<12 & Evaluate [0.5em] [-1em] -3 & 2( -3)^2-5( -3)? <12 & 33 ≮ 12 * [0.5em] [-1em] 0 & 2( 0)^2-5( 0)? <12 & 0 < 12 ✓ [0.5em] [-1em] 5 & 2( 5)^2-5( 5)? <12 & 25 ≮ 12 * [0.5em] As we can see, the inequality is true for values between the boundary points. Therefore, we should shade this region.